Precalculus Polar Form and De Moivre’s Theorem Test Solutions and Scoring (PDF)

Precalculus math test solutions: Polar Form and De Moivre's Theorem – Maths PDF

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This polar complex test answers for Precalculus covers polar form and de moivre's theorem (Common Core N-CN.B.4, N-CN.B.5, N-CN.B.6), with a free PDF to print.

Worked solutions and scoring notes for the 9-question polar form test. Use the scoring line under each solution to award partial credit.

Precalculus · Common Core State Standards: N-CN.B.4, N-CN.B.5, N-CN.B.6

Polar complex test answers: worked solutions

Question 1 – Polar form [2 pts]

\(r=\sqrt{12+4}=4\). Quadrant IV with reference angle \(\arctan\frac{2}{2\sqrt3}=30^{\circ}\), so \(\theta=330^{\circ}\).

\(4(\cos330^{\circ}+i\sin330^{\circ})\)

Scoring: 1 point for r, 1 point for the angle.

Question 2 – Rectangular form [2 pts]

\(8\cos225^{\circ}=-4\sqrt2\) and \(8\sin225^{\circ}=-4\sqrt2\).

\(-4\sqrt2-4\sqrt2\,i\)

Scoring: 1 point for each part.

Question 3 – Product [2 pts]

\(10\,\text{cis}\,120^{\circ}\). Rectangular: \(10\cos120^{\circ}=-5\) and \(10\sin120^{\circ}=5\sqrt3\).

\(10\,\text{cis}\,120^{\circ}=-5+5\sqrt3\,i\)

Scoring: 1 point for polar form, 1 point for rectangular.

Question 4 – Quotient [2 pts]

\(\frac{18}{6}=3\) and \(280^{\circ}-100^{\circ}=180^{\circ}\).

\(3\,\text{cis}\,180^{\circ}=-3\)

Scoring: 1 point for the modulus, 1 point for the angle.

Question 5 – Power with De Moivre [3 pts]

z^6 lies on the positive imaginary axis.

(a) \(r=\sqrt2\), \(\theta=135^{\circ}\). (b) \(z^{6}=(\sqrt2)^{6}\text{cis}(810^{\circ})=8\,\text{cis}\,90^{\circ}\). (c) \(8i\).

\(\sqrt2\,\text{cis}\,135^{\circ}\); \(8\,\text{cis}\,90^{\circ}\); \(8i\)

Scoring: 1 point for each part.

Question 6 – Large power [2 pts]

\(1-i=\sqrt2\,\text{cis}(-45^{\circ})\). \((\sqrt2)^{10}=32\), \(10\cdot(-45^{\circ})=-450^{\circ}\equiv-90^{\circ}\).

\(32\,\text{cis}(-90^{\circ})=-32i\)

Scoring: 1 point for the modulus power, 1 point for the angle and result.

Question 7 – Cube roots [2 pts]

\(-27=27\,\text{cis}\,180^{\circ}\). Roots: \(3\,\text{cis}\,\frac{180^{\circ}+360^{\circ}k}{3}\) for \(k=0,1,2\).

\(3\,\text{cis}\,60^{\circ},\ 3\,\text{cis}\,180^{\circ},\ 3\,\text{cis}\,300^{\circ}\)

Scoring: 1 point for the modulus and method, 1 point for the three angles.

Question 8 – Fourth roots [3 pts]

Roots at 0, 90, 180 and 270 degrees.

(a) \(16\,\text{cis}\,0^{\circ}\). (b) Modulus \(2\), arguments \(0^{\circ},90^{\circ},180^{\circ},270^{\circ}\): \(2,\ 2i,\ -2,\ -2i\). (c) They lie on one circle with equal angles of \(90^{\circ}\) between them.

\(16\,\text{cis}\,0^{\circ}\); \(2,\ 2i,\ -2,\ -2i\); equal spacing of \(90^{\circ}\)

Scoring: 1 point for each part.

Question 9 – Rotation [2 pts]

\(i(-2+5i)=-2i+5i^{2}\).

\(-5-2i\)

Scoring: 1 point for the multiplication, 1 point for simplifying.

Total score

Total

20 points

Key facts behind this polar complex test answers

Key facts

z = a + bi = r(cos theta + i sin theta) with r = sqrt(a^2 + b^2) and tan theta = b/a.
Product: multiply moduli, add arguments. Quotient: divide moduli, subtract arguments.
De Moivre: [r cis theta]^n = r^n cis(n theta).
nth roots: (r)^(1/n) cis((theta + 360 k)/n), k = 0, …, n – 1.
The n roots form a regular n-gon on a circle of radius r^(1/n).

How to use this polar complex test answers

Score your test with the point values shown for each question. Then read the worked solution of every question where you lost points and note the method to review.

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